For PROF XAVIER ONLY

Handout 8

ECO 444

Konrad Grabiszewski

Partial Derivatives

1 Partial Derivatives

Example. Take f(x, y) = x + y, g(x, y) = x2 + 2xy + 3, h(x, y) = y x , and t(x, y) = ex − 2y3x2. We

compute the first- and second-order partial derivatives with respect to x and y. Note that in each

case, we have ∂ 2f

∂x∂y = ∂

2f ∂y∂x

. It is not a coincidence; this is the symmetry of second derivatives and

you might know it as the Young theorem.

∂f

∂x = 1

∂f

∂y = 1

∂2f

∂x2 = 0

∂2f

∂y2 = 0

∂2f

∂x∂y = 0

∂2f

∂y∂x = 0

∂g

∂x = 2x + 2y

∂g

∂y = 2x

∂2g

∂x2 = 2

∂2g

∂y2 = 0

∂2g

∂x∂y = 2

∂2g

∂y∂x = 2

∂h

∂x = −

y

x2

∂h

∂y =

1

x

∂2h

∂x2 = 2

y

x3

∂2h

∂y2 = 0

∂2h

∂x∂y = −

1

x2

∂2h

∂y∂x = −

1

x2

∂t

∂x = ex − 4y3x

∂t

∂y = ex − 6y2x2

∂2t

∂x2 = ex − 4y3

∂2t

∂y2 = −12yx2

∂2t

∂x∂y = ex − 12y2x

∂2t

∂y∂x = ex − 12y2x

1

2 Optimization

Example. For a given y, find the maximum of f(x, y) = −2x2 + 3x + 5y − 3xy with respect to

x. Note that f is defined on R × R. First, we compute the first-order derivative (FOD) and

second-order derivative (SOD).

∂f

∂x = −4x + 3 − 3y (1)

∂2f

∂x2 = −4 (2)

Second, we look at FOD to find a candidate for maximum. Note that if x0 is a local maximum or

minimum, then ∂f ∂x

(x0) = 0. Hence, from FOD we have our candidate x0 = 3y−3 4

. For example, if

y = 0, then x0 = −0.75.

Third, we look at SOD and check whether the second-order condition (SOC) is satisfied. Recall

that a function f attains a maximum at x0 if ∂2f ∂x2

(x0) < 0. Note that we have ∂2f ∂x2

(x) = −4 for

every x ∈ R and every y ∈ R. In particular, ∂ 2f

∂x2

( 3y−3 4

) = −4 < 0 for each y ∈ R. Hence, for a

given y, x0 = 3y−3 4

is a local maximum.

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